LeetCode in Kotlin
1638. Count Substrings That Differ by One Character
Medium
Given two strings s and t, find the number of ways you can choose a non-empty substring of s and replace a single character by a different character such that the resulting substring is a substring of t. In other words, find the number of substrings in s that differ from some substring in t by exactly one character.
For example, the underlined substrings in "computer" and "computation" only differ by the 'e'/'a', so this is a valid way.
Return the number of substrings that satisfy the condition above.
A substring is a contiguous sequence of characters within a string.
Example 1:
Input: s = “aba”, t = “baba”
Output: 6
Explanation: The following are the pairs of substrings from s and t that differ by exactly 1 character:
(“aba”, “baba”)
(“aba”, “baba”)
(“aba”, “baba”)
(“aba”, “baba”)
(“aba”, “baba”)
(“aba”, “baba”)
The underlined portions are the substrings that are chosen from s and t.
Example 2:
Input: s = “ab”, t = “bb”
Output: 3
Explanation: The following are the pairs of substrings from s and t that differ by 1 character:
(“ab”, “bb”)
(“ab”, “bb”)
(“ab”, “bb”)
The underlined portions are the substrings that are chosen from s and t.
Constraints:
1 <= s.length, t.length <= 100sandtconsist of lowercase English letters only.
Solution
class Solution {
fun countSubstrings(s: String, t: String): Int {
var ans = 0
val n = s.length
val m = t.length
val dp = Array(n + 1) { IntArray(m + 1) }
val tp = Array(n + 1) { IntArray(m + 1) }
for (i in n - 1 downTo 0) {
for (j in m - 1 downTo 0) {
if (s[i] == t[j]) {
dp[i][j] = dp[i + 1][j + 1] + 1
tp[i][j] = tp[i + 1][j + 1]
} else {
tp[i][j] = dp[i + 1][j + 1] + 1
}
ans += tp[i][j]
}
}
return ans
}
}